A motor supplies 0.4 hp to stir 5 kg of water. Assuming that...

A motor supplies $0.4$ hp to stir 5 kg of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water $6{ }^{\circ} \mathrm{C}$ ? The heat required to heat the water is $$\Delta Q=m c \Delta T=(5000 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(6{ }^{\circ} \mathrm{C}\right)=30 \mathrm{kcal}$$ This is actually supplied by friction work, so Friction work done $=\Delta Q=(30 \mathrm{kcal})(4.184 \mathrm{~J} / \mathrm{cal})=126 \mathrm{~kJ}$ and this equals the work done by the motor. But Work done by motor in time $t=($ Power $)(t)=(0.4 \mathrm{hp} \times 746 \mathrm{~W} / \mathrm{hp})(t)$. Equating this to our previous value for the work done yields $$ t=\frac{1.26 \times 10^{5} \mathrm{~J}}{(0.4 \times 746) \mathrm{W}}=420 \mathrm{~s}=7 \mathrm{~min} $$

A motor supplies $0.4$ hp to stir 5 kg of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water $6{ }^{\circ} \mathrm{C}$ ?
The heat required to heat the water is
$$\Delta Q=m c \Delta T=(5000 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(6{ }^{\circ} \mathrm{C}\right)=30 \mathrm{kcal}$$
This is actually supplied by friction work, so
Friction work done $=\Delta Q=(30 \mathrm{kcal})(4.184 \mathrm{~J} / \mathrm{cal})=126 \mathrm{~kJ}$
and this equals the work done by the motor. But Work done by motor in time $t=($ Power $)(t)=(0.4 \mathrm{hp} \times 746 \mathrm{~W} / \mathrm{hp})(t)$.
Equating this to our previous value for the work done yields
$$
t=\frac{1.26 \times 10^{5} \mathrm{~J}}{(0.4 \times 746) \mathrm{W}}=420 \mathrm{~s}=7 \mathrm{~min}
$$

Key Concepts

Power

Power refers to the rate at which work is done or energy is transferred. In this problem, power is used to determine how much mechanical work the motor can perform per unit time to generate heat through friction. Converting units, such as from horsepower to watts, is essential to ensure consistency across calculations.

Energy Conversion

Energy conversion is the process of transforming energy from one form to another. Here, the mechanical work provided by the motor is entirely converted into thermal energy via friction, which then heats the water. Recognizing and quantifying this conversion is key to solving problems where work results in temperature changes.

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates the amount of energy required to raise the temperature of a unit mass by one degree Celsius. In the context of this problem, the specific heat capacity of water is used to calculate the total energy needed to achieve the desired temperature increase, linking the energy input to the temperature change.

Unit Conversion

Unit conversion involves converting measurements from one system to another to maintain consistency across calculations, such as converting horsepower to watts or calories to joules. This step is critical in ensuring that all calculations are performed using compatible units, allowing for accurate determination of time and energy.

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